SQ 025

Shared Qs (025)

Question

Match each plot with its function. Different sources define arccot with different ranges. I’ve followed the arccot defined by Desmos and wikipedia, with a range of \(\left[0,\pi\right]\), not \(\left[\frac{-\pi}{2},\,\frac{\pi}{2}\right]\). This choice gives a continuous function, but means that when \(m<0\), \(\arctan(\frac{1}{m})\ne \mathrm{arccot}(m)\). The other possible definition (not used here) is seen on Wolfram Alpha.

Plot 1

Plot 2

Plot 3

Plot 4

Plot 5

Plot 6

Plot 7

Plot 8

Plot 9

Plot 10

Plot 11

Plot 12
Solution

I’d recommend using a graphing utility, like Desmos.
Question

Match each plot with its function. Different sources define arccot with different ranges. I’ve followed the arccot defined by Desmos and wikipedia, with a range of \(\left[0,\pi\right]\), not \(\left[\frac{-\pi}{2},\,\frac{\pi}{2}\right]\). This choice gives a continuous function, but means that when \(m<0\), \(\arctan(\frac{1}{m})\ne \mathrm{arccot}(m)\). The other possible definition (not used here) is seen on Wolfram Alpha.

Plot 1

Plot 2

Plot 3

Plot 4

Plot 5

Plot 6

Plot 7

Plot 8

Plot 9

Plot 10

Plot 11

Plot 12
Solution

I’d recommend using a graphing utility, like Desmos.
Question

Match each plot with its function. Different sources define arccot with different ranges. I’ve followed the arccot defined by Desmos and wikipedia, with a range of \(\left[0,\pi\right]\), not \(\left[\frac{-\pi}{2},\,\frac{\pi}{2}\right]\). This choice gives a continuous function, but means that when \(m<0\), \(\arctan(\frac{1}{m})\ne \mathrm{arccot}(m)\). The other possible definition (not used here) is seen on Wolfram Alpha.

Plot 1

Plot 2

Plot 3

Plot 4

Plot 5

Plot 6

Plot 7

Plot 8

Plot 9

Plot 10

Plot 11

Plot 12
Solution

I’d recommend using a graphing utility, like Desmos.
Question

Match each plot with its function. Different sources define arccot with different ranges. I’ve followed the arccot defined by Desmos and wikipedia, with a range of \(\left[0,\pi\right]\), not \(\left[\frac{-\pi}{2},\,\frac{\pi}{2}\right]\). This choice gives a continuous function, but means that when \(m<0\), \(\arctan(\frac{1}{m})\ne \mathrm{arccot}(m)\). The other possible definition (not used here) is seen on Wolfram Alpha.

Plot 1

Plot 2

Plot 3

Plot 4

Plot 5

Plot 6

Plot 7

Plot 8

Plot 9

Plot 10

Plot 11

Plot 12
Solution

I’d recommend using a graphing utility, like Desmos.
Question

Match each plot with its function. Different sources define arccot with different ranges. I’ve followed the arccot defined by Desmos and wikipedia, with a range of \(\left[0,\pi\right]\), not \(\left[\frac{-\pi}{2},\,\frac{\pi}{2}\right]\). This choice gives a continuous function, but means that when \(m<0\), \(\arctan(\frac{1}{m})\ne \mathrm{arccot}(m)\). The other possible definition (not used here) is seen on Wolfram Alpha.

Plot 1

Plot 2

Plot 3

Plot 4

Plot 5

Plot 6

Plot 7

Plot 8

Plot 9

Plot 10

Plot 11

Plot 12
Solution

I’d recommend using a graphing utility, like Desmos.
Question

Match each plot with its function. Different sources define arccot with different ranges. I’ve followed the arccot defined by Desmos and wikipedia, with a range of \(\left[0,\pi\right]\), not \(\left[\frac{-\pi}{2},\,\frac{\pi}{2}\right]\). This choice gives a continuous function, but means that when \(m<0\), \(\arctan(\frac{1}{m})\ne \mathrm{arccot}(m)\). The other possible definition (not used here) is seen on Wolfram Alpha.

Plot 1

Plot 2

Plot 3

Plot 4

Plot 5

Plot 6

Plot 7

Plot 8

Plot 9

Plot 10

Plot 11

Plot 12
Solution

I’d recommend using a graphing utility, like Desmos.
Question

Match each plot with its function. Different sources define arccot with different ranges. I’ve followed the arccot defined by Desmos and wikipedia, with a range of \(\left[0,\pi\right]\), not \(\left[\frac{-\pi}{2},\,\frac{\pi}{2}\right]\). This choice gives a continuous function, but means that when \(m<0\), \(\arctan(\frac{1}{m})\ne \mathrm{arccot}(m)\). The other possible definition (not used here) is seen on Wolfram Alpha.

Plot 1

Plot 2

Plot 3

Plot 4

Plot 5

Plot 6

Plot 7

Plot 8

Plot 9

Plot 10

Plot 11

Plot 12
Solution

I’d recommend using a graphing utility, like Desmos.
Question

Match each plot with its function. Different sources define arccot with different ranges. I’ve followed the arccot defined by Desmos and wikipedia, with a range of \(\left[0,\pi\right]\), not \(\left[\frac{-\pi}{2},\,\frac{\pi}{2}\right]\). This choice gives a continuous function, but means that when \(m<0\), \(\arctan(\frac{1}{m})\ne \mathrm{arccot}(m)\). The other possible definition (not used here) is seen on Wolfram Alpha.

Plot 1

Plot 2

Plot 3

Plot 4

Plot 5

Plot 6

Plot 7

Plot 8

Plot 9

Plot 10

Plot 11

Plot 12
Solution

I’d recommend using a graphing utility, like Desmos.
Question

Match each plot with its function. Different sources define arccot with different ranges. I’ve followed the arccot defined by Desmos and wikipedia, with a range of \(\left[0,\pi\right]\), not \(\left[\frac{-\pi}{2},\,\frac{\pi}{2}\right]\). This choice gives a continuous function, but means that when \(m<0\), \(\arctan(\frac{1}{m})\ne \mathrm{arccot}(m)\). The other possible definition (not used here) is seen on Wolfram Alpha.

Plot 1

Plot 2

Plot 3

Plot 4

Plot 5

Plot 6

Plot 7

Plot 8

Plot 9

Plot 10

Plot 11

Plot 12
Solution

I’d recommend using a graphing utility, like Desmos.
Question

Match each plot with its function. Different sources define arccot with different ranges. I’ve followed the arccot defined by Desmos and wikipedia, with a range of \(\left[0,\pi\right]\), not \(\left[\frac{-\pi}{2},\,\frac{\pi}{2}\right]\). This choice gives a continuous function, but means that when \(m<0\), \(\arctan(\frac{1}{m})\ne \mathrm{arccot}(m)\). The other possible definition (not used here) is seen on Wolfram Alpha.

Plot 1

Plot 2

Plot 3

Plot 4

Plot 5

Plot 6

Plot 7

Plot 8

Plot 9

Plot 10

Plot 11

Plot 12
Solution

I’d recommend using a graphing utility, like Desmos.
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the amplitude of the sinusoidal function graphed above.

Solution

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(2.9)-(-1.1)}{2} ~=~ \frac{4}{2} ~=~ 2\]

The amplitude is 2.

This exact function could be plotted with the following algebraic definition: \[y(t) = 2 \sin\left(\frac{2\pi}{7.2}(t+2.4)\right)+(0.9)\]

The amplitude is 2; the period is 7.2; the horizontal shift is left by 2.4; and the midline is \(y=0.9\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the amplitude of the sinusoidal function graphed above.

Solution

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(-3.1)-(-4.9)}{2} ~=~ \frac{1.8}{2} ~=~ 0.9\]

The amplitude is 0.9.

This exact function could be plotted with the following algebraic definition: \[y(t) = 0.9 \sin\left(\frac{2\pi}{6.4}(t+2)\right)+(-4)\]

The amplitude is 0.9; the period is 6.4; the horizontal shift is left by 2; and the midline is \(y=-4\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the amplitude of the sinusoidal function graphed above.

Solution

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(3.3)-(-4.7)}{2} ~=~ \frac{8}{2} ~=~ 4\]

The amplitude is 4.

This exact function could be plotted with the following algebraic definition: \[y(t) = 4 \sin\left(\frac{2\pi}{3.2}(t+1.1)\right)+(-0.7)\]

The amplitude is 4; the period is 3.2; the horizontal shift is left by 1.1; and the midline is \(y=-0.7\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the amplitude of the sinusoidal function graphed above.

Solution

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(1.3)-(-4.7)}{2} ~=~ \frac{6}{2} ~=~ 3\]

The amplitude is 3.

This exact function could be plotted with the following algebraic definition: \[y(t) = 3 \sin\left(\frac{2\pi}{6}(t+1.3)\right)+(-1.7)\]

The amplitude is 3; the period is 6; the horizontal shift is left by 1.3; and the midline is \(y=-1.7\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the amplitude of the sinusoidal function graphed above.

Solution

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(4.1)-(-2.5)}{2} ~=~ \frac{6.6}{2} ~=~ 3.3\]

The amplitude is 3.3.

This exact function could be plotted with the following algebraic definition: \[y(t) = 3.3 \sin\left(\frac{2\pi}{6.4}(t+5.4)\right)+(0.8)\]

The amplitude is 3.3; the period is 6.4; the horizontal shift is left by 5.4; and the midline is \(y=0.8\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the amplitude of the sinusoidal function graphed above.

Solution

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(0.8)-(-2.2)}{2} ~=~ \frac{3}{2} ~=~ 1.5\]

The amplitude is 1.5.

This exact function could be plotted with the following algebraic definition: \[y(t) = 1.5 \sin\left(\frac{2\pi}{2.4}(t+0.2)\right)+(-0.7)\]

The amplitude is 1.5; the period is 2.4; the horizontal shift is left by 0.2; and the midline is \(y=-0.7\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the amplitude of the sinusoidal function graphed above.

Solution

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(2.8)-(-0.8)}{2} ~=~ \frac{3.6}{2} ~=~ 1.8\]

The amplitude is 1.8.

This exact function could be plotted with the following algebraic definition: \[y(t) = 1.8 \sin\left(\frac{2\pi}{4.4}(t+4.1)\right)+(1)\]

The amplitude is 1.8; the period is 4.4; the horizontal shift is left by 4.1; and the midline is \(y=1\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the amplitude of the sinusoidal function graphed above.

Solution

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(2.6)-(-4)}{2} ~=~ \frac{6.6}{2} ~=~ 3.3\]

The amplitude is 3.3.

This exact function could be plotted with the following algebraic definition: \[y(t) = 3.3 \sin\left(\frac{2\pi}{4.8}(t+0.8)\right)+(-0.7)\]

The amplitude is 3.3; the period is 4.8; the horizontal shift is left by 0.8; and the midline is \(y=-0.7\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the amplitude of the sinusoidal function graphed above.

Solution

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(3.9)-(-2.3)}{2} ~=~ \frac{6.2}{2} ~=~ 3.1\]

The amplitude is 3.1.

This exact function could be plotted with the following algebraic definition: \[y(t) = 3.1 \sin\left(\frac{2\pi}{4}(t+3.6)\right)+(0.8)\]

The amplitude is 3.1; the period is 4; the horizontal shift is left by 3.6; and the midline is \(y=0.8\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the amplitude of the sinusoidal function graphed above.

Solution

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(3.4)-(-1.4)}{2} ~=~ \frac{4.8}{2} ~=~ 2.4\]

The amplitude is 2.4.

This exact function could be plotted with the following algebraic definition: \[y(t) = 2.4 \sin\left(\frac{2\pi}{6.8}(t+6.4)\right)+(1)\]

The amplitude is 2.4; the period is 6.8; the horizontal shift is left by 6.4; and the midline is \(y=1\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the period of the sinusoidal function graphed above.

Solution

Find the time difference between corresponding points in sequential repetitions.

\[P~=~ (3.8)-(-1)\] \[P = 4.8\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 2.9 \sin\left(\frac{2\pi}{4.8}(t+2.2)\right)+(1.9)\]

The amplitude is 2.9; the period is 4.8; the horizontal shift is left by 2.2; and the midline is \(y=1.9\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the period of the sinusoidal function graphed above.

Solution

Find the time difference between corresponding points in sequential repetitions.

\[P~=~ (2.2)-(-2.6)\] \[P = 4.8\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 2.3 \sin\left(\frac{2\pi}{4.8}(t+3.8)\right)+(-1.2)\]

The amplitude is 2.3; the period is 4.8; the horizontal shift is left by 3.8; and the midline is \(y=-1.2\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the period of the sinusoidal function graphed above.

Solution

Find the time difference between corresponding points in sequential repetitions.

\[P~=~ (-2.7)-(-4.7)\] \[P = 2\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 2.5 \sin\left(\frac{2\pi}{2}(t+1.2)\right)+(-1.3)\]

The amplitude is 2.5; the period is 2; the horizontal shift is left by 1.2; and the midline is \(y=-1.3\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the period of the sinusoidal function graphed above.

Solution

Find the time difference between corresponding points in sequential repetitions.

\[P~=~ (1.5)-(-3.7)\] \[P = 5.2\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 3.3 \sin\left(\frac{2\pi}{5.2}(t+5)\right)+(-1)\]

The amplitude is 3.3; the period is 5.2; the horizontal shift is left by 5; and the midline is \(y=-1\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the period of the sinusoidal function graphed above.

Solution

Find the time difference between corresponding points in sequential repetitions.

\[P~=~ (-0.7)-(-4.3)\] \[P = 3.6\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 2.2 \sin\left(\frac{2\pi}{3.6}(t+1.6)\right)+(2.5)\]

The amplitude is 2.2; the period is 3.6; the horizontal shift is left by 1.6; and the midline is \(y=2.5\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the period of the sinusoidal function graphed above.

Solution

Find the time difference between corresponding points in sequential repetitions.

\[P~=~ (0.2)-(-3)\] \[P = 3.2\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 3.3 \sin\left(\frac{2\pi}{3.2}(t+0.6)\right)+(-0.3)\]

The amplitude is 3.3; the period is 3.2; the horizontal shift is left by 0.6; and the midline is \(y=-0.3\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the period of the sinusoidal function graphed above.

Solution

Find the time difference between corresponding points in sequential repetitions.

\[P~=~ (2.6)-(-3.4)\] \[P = 6\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 3.7 \sin\left(\frac{2\pi}{6}(t+4.9)\right)+(-0.9)\]

The amplitude is 3.7; the period is 6; the horizontal shift is left by 4.9; and the midline is \(y=-0.9\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the period of the sinusoidal function graphed above.

Solution

Find the time difference between corresponding points in sequential repetitions.

\[P~=~ (-1.2)-(-4.8)\] \[P = 3.6\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 1.1 \sin\left(\frac{2\pi}{3.6}(t+2.1)\right)+(-2.7)\]

The amplitude is 1.1; the period is 3.6; the horizontal shift is left by 2.1; and the midline is \(y=-2.7\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the period of the sinusoidal function graphed above.

Solution

Find the time difference between corresponding points in sequential repetitions.

\[P~=~ (0.3)-(-2.9)\] \[P = 3.2\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 2.5 \sin\left(\frac{2\pi}{3.2}(t+0.5)\right)+(-0.3)\]

The amplitude is 2.5; the period is 3.2; the horizontal shift is left by 0.5; and the midline is \(y=-0.3\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the period of the sinusoidal function graphed above.

Solution

Find the time difference between corresponding points in sequential repetitions.

\[P~=~ (0.8)-(-4.4)\] \[P = 5.2\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 0.8 \sin\left(\frac{2\pi}{5.2}(t+0.5)\right)+(-2.7)\]

The amplitude is 0.8; the period is 5.2; the horizontal shift is left by 0.5; and the midline is \(y=-2.7\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the vertical shift of the sinusoidal function graphed above.

Solution

The vertical offset can be found by subtracting the amplitude from the maximum.

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(3.5)-(-0.5)}{2} ~=~ \frac{4}{2} ~=~ 2\]

The amplitude is 2. Subtract 2 from 3.5 (the maximum).

\[D ~=~ 3.5-2\] \[D ~=~ 1.5\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 2 \sin\left(\frac{2\pi}{2.8}(t+1.8)\right)+(1.5)\]

The amplitude is 2; the period is 2.8; the horizontal shift is left by 1.8; and the midline is \(y=1.5\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the vertical shift of the sinusoidal function graphed above.

Solution

The vertical offset can be found by subtracting the amplitude from the maximum.

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(0.8)-(-4.6)}{2} ~=~ \frac{5.4}{2} ~=~ 2.7\]

The amplitude is 2.7. Subtract 2.7 from 0.8 (the maximum).

\[D ~=~ 0.8-2.7\] \[D ~=~ -1.9\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 2.7 \sin\left(\frac{2\pi}{6.8}(t+3.3)\right)+(-1.9)\]

The amplitude is 2.7; the period is 6.8; the horizontal shift is left by 3.3; and the midline is \(y=-1.9\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the vertical shift of the sinusoidal function graphed above.

Solution

The vertical offset can be found by subtracting the amplitude from the maximum.

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(2.1)-(-4.7)}{2} ~=~ \frac{6.8}{2} ~=~ 3.4\]

The amplitude is 3.4. Subtract 3.4 from 2.1 (the maximum).

\[D ~=~ 2.1-3.4\] \[D ~=~ -1.3\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 3.4 \sin\left(\frac{2\pi}{8}(t+3.7)\right)+(-1.3)\]

The amplitude is 3.4; the period is 8; the horizontal shift is left by 3.7; and the midline is \(y=-1.3\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the vertical shift of the sinusoidal function graphed above.

Solution

The vertical offset can be found by subtracting the amplitude from the maximum.

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(-1.3)-(-4.7)}{2} ~=~ \frac{3.4}{2} ~=~ 1.7\]

The amplitude is 1.7. Subtract 1.7 from -1.3 (the maximum).

\[D ~=~ -1.3-1.7\] \[D ~=~ -3\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 1.7 \sin\left(\frac{2\pi}{4.4}(t+3.3)\right)+(-3)\]

The amplitude is 1.7; the period is 4.4; the horizontal shift is left by 3.3; and the midline is \(y=-3\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the vertical shift of the sinusoidal function graphed above.

Solution

The vertical offset can be found by subtracting the amplitude from the maximum.

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(3.1)-(-2.1)}{2} ~=~ \frac{5.2}{2} ~=~ 2.6\]

The amplitude is 2.6. Subtract 2.6 from 3.1 (the maximum).

\[D ~=~ 3.1-2.6\] \[D ~=~ 0.5\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 2.6 \sin\left(\frac{2\pi}{5.6}(t+2.2)\right)+(0.5)\]

The amplitude is 2.6; the period is 5.6; the horizontal shift is left by 2.2; and the midline is \(y=0.5\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the vertical shift of the sinusoidal function graphed above.

Solution

The vertical offset can be found by subtracting the amplitude from the maximum.

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(4.3)-(1.9)}{2} ~=~ \frac{2.4}{2} ~=~ 1.2\]

The amplitude is 1.2. Subtract 1.2 from 4.3 (the maximum).

\[D ~=~ 4.3-1.2\] \[D ~=~ 3.1\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 1.2 \sin\left(\frac{2\pi}{6}(t+1.4)\right)+(3.1)\]

The amplitude is 1.2; the period is 6; the horizontal shift is left by 1.4; and the midline is \(y=3.1\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the vertical shift of the sinusoidal function graphed above.

Solution

The vertical offset can be found by subtracting the amplitude from the maximum.

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(1.7)-(-1.9)}{2} ~=~ \frac{3.6}{2} ~=~ 1.8\]

The amplitude is 1.8. Subtract 1.8 from 1.7 (the maximum).

\[D ~=~ 1.7-1.8\] \[D ~=~ -0.1\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 1.8 \sin\left(\frac{2\pi}{7.2}(t+0.1)\right)+(-0.1)\]

The amplitude is 1.8; the period is 7.2; the horizontal shift is left by 0.1; and the midline is \(y=-0.1\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the vertical shift of the sinusoidal function graphed above.

Solution

The vertical offset can be found by subtracting the amplitude from the maximum.

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(2.4)-(-4.2)}{2} ~=~ \frac{6.6}{2} ~=~ 3.3\]

The amplitude is 3.3. Subtract 3.3 from 2.4 (the maximum).

\[D ~=~ 2.4-3.3\] \[D ~=~ -0.9\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 3.3 \sin\left(\frac{2\pi}{5.2}(t+2)\right)+(-0.9)\]

The amplitude is 3.3; the period is 5.2; the horizontal shift is left by 2; and the midline is \(y=-0.9\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the vertical shift of the sinusoidal function graphed above.

Solution

The vertical offset can be found by subtracting the amplitude from the maximum.

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(2.2)-(0.4)}{2} ~=~ \frac{1.8}{2} ~=~ 0.9\]

The amplitude is 0.9. Subtract 0.9 from 2.2 (the maximum).

\[D ~=~ 2.2-0.9\] \[D ~=~ 1.3\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 0.9 \sin\left(\frac{2\pi}{5.2}(t+1.3)\right)+(1.3)\]

The amplitude is 0.9; the period is 5.2; the horizontal shift is left by 1.3; and the midline is \(y=1.3\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the vertical shift of the sinusoidal function graphed above.

Solution

The vertical offset can be found by subtracting the amplitude from the maximum.

The amplitude of a sinusoidal function is half the difference between the maximum \(y\) and the minimum \(y\).

\[A~=~\frac{(1.9)-(-4.9)}{2} ~=~ \frac{6.8}{2} ~=~ 3.4\]

The amplitude is 3.4. Subtract 3.4 from 1.9 (the maximum).

\[D ~=~ 1.9-3.4\] \[D ~=~ -1.5\]

This exact function could be plotted with the following algebraic definition: \[y(t) = 3.4 \sin\left(\frac{2\pi}{4.4}(t+3.1)\right)+(-1.5)\]

The amplitude is 3.4; the period is 4.4; the horizontal shift is left by 3.1; and the midline is \(y=-1.5\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the leftward (back in time) shift of the sinusoidal function graphed above.

Solution

Determine the midline. Find the point on the midline that is to the left of the origin and has positive slope. This key point is mapped from the origin of the parent sine function.

The leftward shift is \(2.5\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the leftward (back in time) shift of the sinusoidal function graphed above.

Solution

Determine the midline. Find the point on the midline that is to the left of the origin and has positive slope. This key point is mapped from the origin of the parent sine function.

The leftward shift is \(1.5\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the leftward (back in time) shift of the sinusoidal function graphed above.

Solution

Determine the midline. Find the point on the midline that is to the left of the origin and has positive slope. This key point is mapped from the origin of the parent sine function.

The leftward shift is \(0.3\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the leftward (back in time) shift of the sinusoidal function graphed above.

Solution

Determine the midline. Find the point on the midline that is to the left of the origin and has positive slope. This key point is mapped from the origin of the parent sine function.

The leftward shift is \(2.5\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the leftward (back in time) shift of the sinusoidal function graphed above.

Solution

Determine the midline. Find the point on the midline that is to the left of the origin and has positive slope. This key point is mapped from the origin of the parent sine function.

The leftward shift is \(3.9\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the leftward (back in time) shift of the sinusoidal function graphed above.

Solution

Determine the midline. Find the point on the midline that is to the left of the origin and has positive slope. This key point is mapped from the origin of the parent sine function.

The leftward shift is \(2.9\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the leftward (back in time) shift of the sinusoidal function graphed above.

Solution

Determine the midline. Find the point on the midline that is to the left of the origin and has positive slope. This key point is mapped from the origin of the parent sine function.

The leftward shift is \(4.1\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the leftward (back in time) shift of the sinusoidal function graphed above.

Solution

Determine the midline. Find the point on the midline that is to the left of the origin and has positive slope. This key point is mapped from the origin of the parent sine function.

The leftward shift is \(0.7\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the leftward (back in time) shift of the sinusoidal function graphed above.

Solution

Determine the midline. Find the point on the midline that is to the left of the origin and has positive slope. This key point is mapped from the origin of the parent sine function.

The leftward shift is \(1.4\).
Question

Some quantity (\(y\)) is varying in time (\(t\)), following a temporal sinusoidal function parameterized with 4 parameters:

\[y(t) ~=~ A\sin\left(\frac{2\pi}{P}(t+L)\right)+D\]
- Amplitude \(A\), where \(A>0\)
  - Amplitude is the “vertical stretch” factor.
  - A negative \(A\) would cause a vertical reflection. However, an equivalent transformation would be caused by shifting “horizontally” (in time) by half a period, so there is no loss of generality by forcing \(A\) to be positive.
  - The amplitude might indicate changes in pressure, position, temperature, or many other types of quantities.
  - The wikipedia article on amplitude refers to \(A\) as the semi-amplitude.
- Period \(P\), where \(P>0\)
  - Period is a “horizontal stretch” factor.
  - Another symbol, \(T\), is often used for period.
  - Period has units of time, typically seconds.
  - The frequency is the reciprocal of \(P\), so \(f=\frac{1}{P}\). Frequency has units of hertz.
  - The angular frequency \(\omega\) is in radians per second: \(\omega~=~\frac{2\pi}{P}~=~2\pi f\)
  - If the independent variable is spatial rather than temporal, the analogous term to “period” is “wavelength”; often \(\lambda\) is used for wavelength.
  - Most physical waves (like sound and light) vary in space and time, so they have both wavelength and period, which are related by the speed (\(v\)) of the wave: \(v=\frac{\lambda}{P}\)
- Leftward shift \(L\), where \(0\le L<P\). All points along the sine function are shifted backwards (in time) by \(L\).
  - When \(t=0\), it is kind of like \(L\) time has already passed (compared to the parent sine function).
  - Like \(A\), the parameter \(L\) does not need to be constrained (between 0 and \(P\)), but any sinusoidal wave can be generated with this constraint.
  - This parameter, \(L\), has units of time.
  - Many textbooks will incorrectly/confusingly refer to this as the “phase shift”.
  - The phase shift, \(\phi\), has units of radians: \(\phi=\frac{2\pi L}{P}\).
- Vertical shift \(D\)
  - This vertical shift is called DC offset or DC bias in sound recording.
  - The midline at \(y=D\) is halfway between the crests and troughs.
Those four parameters were randomly assigned values, producing the graph below.

Estimate the leftward (back in time) shift of the sinusoidal function graphed above.

Solution

Determine the midline. Find the point on the midline that is to the left of the origin and has positive slope. This key point is mapped from the origin of the parent sine function.

The leftward shift is \(1\).
Question

To generate the plot below, four parameters (\(A\), \(B\), \(C\), and \(D\)) where randomly selected and plugged into the following sinusoidal function:

\[y(t) = A\sin(Bt+C)+D\]

where \[A>0\] and \[0\le C < 2 \pi\]

The first maximum is \((0.06, 1.17)\) and the first minimum is \((0.28, 0.29)\). Determine the values of the parameters (rounded to the nearest hundredth).
- \(A =\)
- \(B =\)
- \(C =\)
- \(D =\)
Solution
- \(A = 0.44\)
- \(B = 14.28\)
- \(C = 0.71\)
- \(D = 0.73\)
The easiest parameters to find are \(A\) and \(D\). With the given parameterization, \(A\) is the (semi-)amplitude and \(D\) is the vertical shift of the midline.

\[A ~=~ \frac{(1.17)-(0.29)}{2} ~=~ 0.44\]

\[D ~=~ \frac{(1.17)+(0.29)}{2} ~=~ 0.73\]

In the given parameterization, \(B\) represents the angular frequency. It is inversely proportional to the period, which is easy to determine from the 2 given points. The two marked points are half a period away from each other (in time). \[P ~=~ 2\cdot(0.28-0.06) ~=~ 0.44\]

You can calculate the angular frequency by dividing \(2\pi\) by the period.

\[B = \frac{2\pi}{0.44} = 14.2799666\] We round to the hundredths place.

\[B \approx 14.28\]

With the given parameterization, \(C\) represents the phase (in radians), representing how far into the cycle the wave is when \(t=0\). It will be proportional to the horizontal shift. The horizontal shift is to the left, because a positive \(C\) is added inside the argument. The parent function (sine) has a point at the origin with a positive slope. Find the analogous point (on the midline with positive slope) to the left of the vertical axis.

First, determine the quarter period is \(P/4 = 0.11\). Then, recognize that the key point is 1 quarter periods before the first given point. So, to find the \(t\) value of the key point: \[0.06-1\cdot0.11 ~=~ -0.05\] And so the key point is at \((-0.05, 0.73)\).

The sine function was shifted 0.05 time units to the left. If we divide 0.05 by the period and multiply by \(2\pi\), it tells us the phase.

\[C = \frac{0.05}{0.44}\cdot 2\pi = 0.7139983\]

Round to the nearest hundredth:

\[C \approx 0.71\]
Question

To generate the plot below, four parameters (\(A\), \(B\), \(C\), and \(D\)) where randomly selected and plugged into the following sinusoidal function:

\[y(t) = A\sin(Bt+C)+D\]

where \[A>0\] and \[0\le C < 2 \pi\]

The first maximum is \((2.42, 1.54)\) and the first minimum is \((1.16, -1.06)\). Determine the values of the parameters (rounded to the nearest hundredth).
- \(A =\)
- \(B =\)
- \(C =\)
- \(D =\)
Solution
- \(A = 1.30\)
- \(B = 2.49\)
- \(C = 1.82\)
- \(D = 0.24\)
The easiest parameters to find are \(A\) and \(D\). With the given parameterization, \(A\) is the (semi-)amplitude and \(D\) is the vertical shift of the midline.

\[A ~=~ \frac{(1.54)-(-1.06)}{2} ~=~ 1.3\]

\[D ~=~ \frac{(1.54)+(-1.06)}{2} ~=~ 0.24\]

In the given parameterization, \(B\) represents the angular frequency. It is inversely proportional to the period, which is easy to determine from the 2 given points. The two marked points are half a period away from each other (in time). \[P ~=~ 2\cdot(2.42-1.16) ~=~ 2.52\]

You can calculate the angular frequency by dividing \(2\pi\) by the period.

\[B = \frac{2\pi}{2.52} = 2.4933275\] We round to the hundredths place.

\[B \approx 2.49\]

With the given parameterization, \(C\) represents the phase (in radians), representing how far into the cycle the wave is when \(t=0\). It will be proportional to the horizontal shift. The horizontal shift is to the left, because a positive \(C\) is added inside the argument. The parent function (sine) has a point at the origin with a positive slope. Find the analogous point (on the midline with positive slope) to the left of the vertical axis.

First, determine the quarter period is \(P/4 = 0.63\). Then, recognize that the key point is 3 quarter periods before the first given point. So, to find the \(t\) value of the key point: \[1.16-3\cdot0.63 ~=~ -0.73\] And so the key point is at \((-0.73, 0.24)\).

The sine function was shifted 0.73 time units to the left. If we divide 0.73 by the period and multiply by \(2\pi\), it tells us the phase.

\[C = \frac{0.73}{2.52}\cdot 2\pi = 1.8201291\]

Round to the nearest hundredth:

\[C \approx 1.82\]
Question

To generate the plot below, four parameters (\(A\), \(B\), \(C\), and \(D\)) where randomly selected and plugged into the following sinusoidal function:

\[y(t) = A\sin(Bt+C)+D\]

where \[A>0\] and \[0\le C < 2 \pi\]

The first maximum is \((1.39, 1.01)\) and the first minimum is \((0.45, -1.61)\). Determine the values of the parameters (rounded to the nearest hundredth).
- \(A =\)
- \(B =\)
- \(C =\)
- \(D =\)
Solution
- \(A = 1.31\)
- \(B = 3.34\)
- \(C = 3.21\)
- \(D = -0.30\)
The easiest parameters to find are \(A\) and \(D\). With the given parameterization, \(A\) is the (semi-)amplitude and \(D\) is the vertical shift of the midline.

\[A ~=~ \frac{(1.01)-(-1.61)}{2} ~=~ 1.31\]

\[D ~=~ \frac{(1.01)+(-1.61)}{2} ~=~ -0.3\]

In the given parameterization, \(B\) represents the angular frequency. It is inversely proportional to the period, which is easy to determine from the 2 given points. The two marked points are half a period away from each other (in time). \[P ~=~ 2\cdot(1.39-0.45) ~=~ 1.88\]

You can calculate the angular frequency by dividing \(2\pi\) by the period.

\[B = \frac{2\pi}{1.88} = 3.3421198\] We round to the hundredths place.

\[B \approx 3.34\]

With the given parameterization, \(C\) represents the phase (in radians), representing how far into the cycle the wave is when \(t=0\). It will be proportional to the horizontal shift. The horizontal shift is to the left, because a positive \(C\) is added inside the argument. The parent function (sine) has a point at the origin with a positive slope. Find the analogous point (on the midline with positive slope) to the left of the vertical axis.

First, determine the quarter period is \(P/4 = 0.47\). Then, recognize that the key point is 3 quarter periods before the first given point. So, to find the \(t\) value of the key point: \[0.45-3\cdot0.47 ~=~ -0.96\] And so the key point is at \((-0.96, -0.3)\).

The sine function was shifted 0.96 time units to the left. If we divide 0.96 by the period and multiply by \(2\pi\), it tells us the phase.

\[C = \frac{0.96}{1.88}\cdot 2\pi = 3.2084351\]

Round to the nearest hundredth:

\[C \approx 3.21\]
Question

To generate the plot below, four parameters (\(A\), \(B\), \(C\), and \(D\)) where randomly selected and plugged into the following sinusoidal function:

\[y(t) = A\sin(Bt+C)+D\]

where \[A>0\] and \[0\le C < 2 \pi\]

The first maximum is \((0.98, 1.09)\) and the first minimum is \((0.46, 0.41)\). Determine the values of the parameters (rounded to the nearest hundredth).
- \(A =\)
- \(B =\)
- \(C =\)
- \(D =\)
Solution
- \(A = 0.34\)
- \(B = 6.04\)
- \(C = 1.93\)
- \(D = 0.75\)
The easiest parameters to find are \(A\) and \(D\). With the given parameterization, \(A\) is the (semi-)amplitude and \(D\) is the vertical shift of the midline.

\[A ~=~ \frac{(1.09)-(0.41)}{2} ~=~ 0.34\]

\[D ~=~ \frac{(1.09)+(0.41)}{2} ~=~ 0.75\]

In the given parameterization, \(B\) represents the angular frequency. It is inversely proportional to the period, which is easy to determine from the 2 given points. The two marked points are half a period away from each other (in time). \[P ~=~ 2\cdot(0.98-0.46) ~=~ 1.04\]

You can calculate the angular frequency by dividing \(2\pi\) by the period.

\[B = \frac{2\pi}{1.04} = 6.0415243\] We round to the hundredths place.

\[B \approx 6.04\]

With the given parameterization, \(C\) represents the phase (in radians), representing how far into the cycle the wave is when \(t=0\). It will be proportional to the horizontal shift. The horizontal shift is to the left, because a positive \(C\) is added inside the argument. The parent function (sine) has a point at the origin with a positive slope. Find the analogous point (on the midline with positive slope) to the left of the vertical axis.

First, determine the quarter period is \(P/4 = 0.26\). Then, recognize that the key point is 3 quarter periods before the first given point. So, to find the \(t\) value of the key point: \[0.46-3\cdot0.26 ~=~ -0.32\] And so the key point is at \((-0.32, 0.75)\).

The sine function was shifted 0.32 time units to the left. If we divide 0.32 by the period and multiply by \(2\pi\), it tells us the phase.

\[C = \frac{0.32}{1.04}\cdot 2\pi = 1.9332878\]

Round to the nearest hundredth:

\[C \approx 1.93\]
Question

To generate the plot below, four parameters (\(A\), \(B\), \(C\), and \(D\)) where randomly selected and plugged into the following sinusoidal function:

\[y(t) = A\sin(Bt+C)+D\]

where \[A>0\] and \[0\le C < 2 \pi\]

The first maximum is \((2.64, 1.24)\) and the first minimum is \((1.16, 0.18)\). Determine the values of the parameters (rounded to the nearest hundredth).
- \(A =\)
- \(B =\)
- \(C =\)
- \(D =\)
Solution
- \(A = 0.53\)
- \(B = 2.12\)
- \(C = 2.25\)
- \(D = 0.71\)
The easiest parameters to find are \(A\) and \(D\). With the given parameterization, \(A\) is the (semi-)amplitude and \(D\) is the vertical shift of the midline.

\[A ~=~ \frac{(1.24)-(0.18)}{2} ~=~ 0.53\]

\[D ~=~ \frac{(1.24)+(0.18)}{2} ~=~ 0.71\]

In the given parameterization, \(B\) represents the angular frequency. It is inversely proportional to the period, which is easy to determine from the 2 given points. The two marked points are half a period away from each other (in time). \[P ~=~ 2\cdot(2.64-1.16) ~=~ 2.96\]

You can calculate the angular frequency by dividing \(2\pi\) by the period.

\[B = \frac{2\pi}{2.96} = 2.1226977\] We round to the hundredths place.

\[B \approx 2.12\]

With the given parameterization, \(C\) represents the phase (in radians), representing how far into the cycle the wave is when \(t=0\). It will be proportional to the horizontal shift. The horizontal shift is to the left, because a positive \(C\) is added inside the argument. The parent function (sine) has a point at the origin with a positive slope. Find the analogous point (on the midline with positive slope) to the left of the vertical axis.

First, determine the quarter period is \(P/4 = 0.74\). Then, recognize that the key point is 3 quarter periods before the first given point. So, to find the \(t\) value of the key point: \[1.16-3\cdot0.74 ~=~ -1.06\] And so the key point is at \((-1.06, 0.71)\).

The sine function was shifted 1.06 time units to the left. If we divide 1.06 by the period and multiply by \(2\pi\), it tells us the phase.

\[C = \frac{1.06}{2.96}\cdot 2\pi = 2.2500596\]

Round to the nearest hundredth:

\[C \approx 2.25\]
Question

To generate the plot below, four parameters (\(A\), \(B\), \(C\), and \(D\)) where randomly selected and plugged into the following sinusoidal function:

\[y(t) = A\sin(Bt+C)+D\]

where \[A>0\] and \[0\le C < 2 \pi\]

The first maximum is \((0.01, 0.57)\) and the first minimum is \((0.19, -1.79)\). Determine the values of the parameters (rounded to the nearest hundredth).
- \(A =\)
- \(B =\)
- \(C =\)
- \(D =\)
Solution
- \(A = 1.18\)
- \(B = 17.45\)
- \(C = 1.40\)
- \(D = -0.61\)
The easiest parameters to find are \(A\) and \(D\). With the given parameterization, \(A\) is the (semi-)amplitude and \(D\) is the vertical shift of the midline.

\[A ~=~ \frac{(0.57)-(-1.79)}{2} ~=~ 1.18\]

\[D ~=~ \frac{(0.57)+(-1.79)}{2} ~=~ -0.61\]

In the given parameterization, \(B\) represents the angular frequency. It is inversely proportional to the period, which is easy to determine from the 2 given points. The two marked points are half a period away from each other (in time). \[P ~=~ 2\cdot(0.19-0.01) ~=~ 0.36\]

You can calculate the angular frequency by dividing \(2\pi\) by the period.

\[B = \frac{2\pi}{0.36} = 17.4532925\] We round to the hundredths place.

\[B \approx 17.45\]

With the given parameterization, \(C\) represents the phase (in radians), representing how far into the cycle the wave is when \(t=0\). It will be proportional to the horizontal shift. The horizontal shift is to the left, because a positive \(C\) is added inside the argument. The parent function (sine) has a point at the origin with a positive slope. Find the analogous point (on the midline with positive slope) to the left of the vertical axis.

First, determine the quarter period is \(P/4 = 0.09\). Then, recognize that the key point is 1 quarter periods before the first given point. So, to find the \(t\) value of the key point: \[0.01-1\cdot0.09 ~=~ -0.08\] And so the key point is at \((-0.08, -0.61)\).

The sine function was shifted 0.08 time units to the left. If we divide 0.08 by the period and multiply by \(2\pi\), it tells us the phase.

\[C = \frac{0.08}{0.36}\cdot 2\pi = 1.3962634\]

Round to the nearest hundredth:

\[C \approx 1.4\]
Question

To generate the plot below, four parameters (\(A\), \(B\), \(C\), and \(D\)) where randomly selected and plugged into the following sinusoidal function:

\[y(t) = A\sin(Bt+C)+D\]

where \[A>0\] and \[0\le C < 2 \pi\]

The first maximum is \((0.46, 0.5)\) and the first minimum is \((0.22, -0.94)\). Determine the values of the parameters (rounded to the nearest hundredth).
- \(A =\)
- \(B =\)
- \(C =\)
- \(D =\)
Solution
- \(A = 0.72\)
- \(B = 13.09\)
- \(C = 1.83\)
- \(D = -0.22\)
The easiest parameters to find are \(A\) and \(D\). With the given parameterization, \(A\) is the (semi-)amplitude and \(D\) is the vertical shift of the midline.

\[A ~=~ \frac{(0.5)-(-0.94)}{2} ~=~ 0.72\]

\[D ~=~ \frac{(0.5)+(-0.94)}{2} ~=~ -0.22\]

In the given parameterization, \(B\) represents the angular frequency. It is inversely proportional to the period, which is easy to determine from the 2 given points. The two marked points are half a period away from each other (in time). \[P ~=~ 2\cdot(0.46-0.22) ~=~ 0.48\]

You can calculate the angular frequency by dividing \(2\pi\) by the period.

\[B = \frac{2\pi}{0.48} = 13.0899694\] We round to the hundredths place.

\[B \approx 13.09\]

With the given parameterization, \(C\) represents the phase (in radians), representing how far into the cycle the wave is when \(t=0\). It will be proportional to the horizontal shift. The horizontal shift is to the left, because a positive \(C\) is added inside the argument. The parent function (sine) has a point at the origin with a positive slope. Find the analogous point (on the midline with positive slope) to the left of the vertical axis.

First, determine the quarter period is \(P/4 = 0.12\). Then, recognize that the key point is 3 quarter periods before the first given point. So, to find the \(t\) value of the key point: \[0.22-3\cdot0.12 ~=~ -0.14\] And so the key point is at \((-0.14, -0.22)\).

The sine function was shifted 0.14 time units to the left. If we divide 0.14 by the period and multiply by \(2\pi\), it tells us the phase.

\[C = \frac{0.14}{0.48}\cdot 2\pi = 1.8325957\]

Round to the nearest hundredth:

\[C \approx 1.83\]
Question

To generate the plot below, four parameters (\(A\), \(B\), \(C\), and \(D\)) where randomly selected and plugged into the following sinusoidal function:

\[y(t) = A\sin(Bt+C)+D\]

where \[A>0\] and \[0\le C < 2 \pi\]

The first maximum is \((0.22, 2.11)\) and the first minimum is \((0.02, -2.65)\). Determine the values of the parameters (rounded to the nearest hundredth).
- \(A =\)
- \(B =\)
- \(C =\)
- \(D =\)
Solution
- \(A = 2.38\)
- \(B = 15.71\)
- \(C = 4.40\)
- \(D = -0.27\)
The easiest parameters to find are \(A\) and \(D\). With the given parameterization, \(A\) is the (semi-)amplitude and \(D\) is the vertical shift of the midline.

\[A ~=~ \frac{(2.11)-(-2.65)}{2} ~=~ 2.38\]

\[D ~=~ \frac{(2.11)+(-2.65)}{2} ~=~ -0.27\]

In the given parameterization, \(B\) represents the angular frequency. It is inversely proportional to the period, which is easy to determine from the 2 given points. The two marked points are half a period away from each other (in time). \[P ~=~ 2\cdot(0.22-0.02) ~=~ 0.4\]

You can calculate the angular frequency by dividing \(2\pi\) by the period.

\[B = \frac{2\pi}{0.4} = 15.7079633\] We round to the hundredths place.

\[B \approx 15.71\]

With the given parameterization, \(C\) represents the phase (in radians), representing how far into the cycle the wave is when \(t=0\). It will be proportional to the horizontal shift. The horizontal shift is to the left, because a positive \(C\) is added inside the argument. The parent function (sine) has a point at the origin with a positive slope. Find the analogous point (on the midline with positive slope) to the left of the vertical axis.

First, determine the quarter period is \(P/4 = 0.1\). Then, recognize that the key point is 3 quarter periods before the first given point. So, to find the \(t\) value of the key point: \[0.02-3\cdot0.1 ~=~ -0.28\] And so the key point is at \((-0.28, -0.27)\).

The sine function was shifted 0.28 time units to the left. If we divide 0.28 by the period and multiply by \(2\pi\), it tells us the phase.

\[C = \frac{0.28}{0.4}\cdot 2\pi = 4.3982297\]

Round to the nearest hundredth:

\[C \approx 4.4\]
Question

To generate the plot below, four parameters (\(A\), \(B\), \(C\), and \(D\)) where randomly selected and plugged into the following sinusoidal function:

\[y(t) = A\sin(Bt+C)+D\]

where \[A>0\] and \[0\le C < 2 \pi\]

The first maximum is \((1.75, 1.33)\) and the first minimum is \((0.53, 0.27)\). Determine the values of the parameters (rounded to the nearest hundredth).
- \(A =\)
- \(B =\)
- \(C =\)
- \(D =\)
Solution
- \(A = 0.53\)
- \(B = 2.58\)
- \(C = 3.35\)
- \(D = 0.80\)
The easiest parameters to find are \(A\) and \(D\). With the given parameterization, \(A\) is the (semi-)amplitude and \(D\) is the vertical shift of the midline.

\[A ~=~ \frac{(1.33)-(0.27)}{2} ~=~ 0.53\]

\[D ~=~ \frac{(1.33)+(0.27)}{2} ~=~ 0.8\]

In the given parameterization, \(B\) represents the angular frequency. It is inversely proportional to the period, which is easy to determine from the 2 given points. The two marked points are half a period away from each other (in time). \[P ~=~ 2\cdot(1.75-0.53) ~=~ 2.44\]

You can calculate the angular frequency by dividing \(2\pi\) by the period.

\[B = \frac{2\pi}{2.44} = 2.5750759\] We round to the hundredths place.

\[B \approx 2.58\]

With the given parameterization, \(C\) represents the phase (in radians), representing how far into the cycle the wave is when \(t=0\). It will be proportional to the horizontal shift. The horizontal shift is to the left, because a positive \(C\) is added inside the argument. The parent function (sine) has a point at the origin with a positive slope. Find the analogous point (on the midline with positive slope) to the left of the vertical axis.

First, determine the quarter period is \(P/4 = 0.61\). Then, recognize that the key point is 3 quarter periods before the first given point. So, to find the \(t\) value of the key point: \[0.53-3\cdot0.61 ~=~ -1.3\] And so the key point is at \((-1.3, 0.8)\).

The sine function was shifted 1.3 time units to the left. If we divide 1.3 by the period and multiply by \(2\pi\), it tells us the phase.

\[C = \frac{1.3}{2.44}\cdot 2\pi = 3.3475987\]

Round to the nearest hundredth:

\[C \approx 3.35\]
Question

To generate the plot below, four parameters (\(A\), \(B\), \(C\), and \(D\)) where randomly selected and plugged into the following sinusoidal function:

\[y(t) = A\sin(Bt+C)+D\]

where \[A>0\] and \[0\le C < 2 \pi\]

The first maximum is \((0.33, 0.79)\) and the first minimum is \((0.81, -1.91)\). Determine the values of the parameters (rounded to the nearest hundredth).
- \(A =\)
- \(B =\)
- \(C =\)
- \(D =\)
Solution
- \(A = 1.35\)
- \(B = 6.54\)
- \(C = 5.69\)
- \(D = -0.56\)
The easiest parameters to find are \(A\) and \(D\). With the given parameterization, \(A\) is the (semi-)amplitude and \(D\) is the vertical shift of the midline.

\[A ~=~ \frac{(0.79)-(-1.91)}{2} ~=~ 1.35\]

\[D ~=~ \frac{(0.79)+(-1.91)}{2} ~=~ -0.56\]

In the given parameterization, \(B\) represents the angular frequency. It is inversely proportional to the period, which is easy to determine from the 2 given points. The two marked points are half a period away from each other (in time). \[P ~=~ 2\cdot(0.81-0.33) ~=~ 0.96\]

You can calculate the angular frequency by dividing \(2\pi\) by the period.

\[B = \frac{2\pi}{0.96} = 6.5449847\] We round to the hundredths place.

\[B \approx 6.54\]

With the given parameterization, \(C\) represents the phase (in radians), representing how far into the cycle the wave is when \(t=0\). It will be proportional to the horizontal shift. The horizontal shift is to the left, because a positive \(C\) is added inside the argument. The parent function (sine) has a point at the origin with a positive slope. Find the analogous point (on the midline with positive slope) to the left of the vertical axis.

First, determine the quarter period is \(P/4 = 0.24\). Then, recognize that the key point is 5 quarter periods before the first given point. So, to find the \(t\) value of the key point: \[0.33-5\cdot0.24 ~=~ -0.87\] And so the key point is at \((-0.87, -0.56)\).

The sine function was shifted 0.87 time units to the left. If we divide 0.87 by the period and multiply by \(2\pi\), it tells us the phase.

\[C = \frac{0.87}{0.96}\cdot 2\pi = 5.6941367\]

Round to the nearest hundredth:

\[C \approx 5.69\]
Question

A sinusoidal pure tone is generated by randomly choosing a frequency (\(F\)) in the following equation: \[y(t) ~=~ \cos(2\pi F t)\]

As shown above, the curve has maximums at both \(t=0\) and \(t=0.94\). What is the frequency in Hz? The tolerance is \(\pm 0.01\) Hz.

Solution

Count how many periods occur between \(t=0\) and \(t=0.94\).

Find the period.

\[P = \frac{0.94 ~\mathrm{seconds}}{6~\mathrm{cycles}}\]

The frequency is the reciprocal of the period.

\[F ~=~ \frac{1}{P} ~=~ \frac{6~\mathrm{cycles}}{0.94 ~\mathrm{seconds}}\]

\[F ~\approx~ 6.38 ~\mathrm{Hz}\]
Question

A sinusoidal pure tone is generated by randomly choosing a frequency (\(F\)) in the following equation: \[y(t) ~=~ \cos(2\pi F t)\]

As shown above, the curve has maximums at both \(t=0\) and \(t=0.9\). What is the frequency in Hz? The tolerance is \(\pm 0.01\) Hz.

Solution

Count how many periods occur between \(t=0\) and \(t=0.9\).

Find the period.

\[P = \frac{0.9 ~\mathrm{seconds}}{4~\mathrm{cycles}}\]

The frequency is the reciprocal of the period.

\[F ~=~ \frac{1}{P} ~=~ \frac{4~\mathrm{cycles}}{0.9 ~\mathrm{seconds}}\]

\[F ~\approx~ 4.44 ~\mathrm{Hz}\]
Question

A sinusoidal pure tone is generated by randomly choosing a frequency (\(F\)) in the following equation: \[y(t) ~=~ \cos(2\pi F t)\]

As shown above, the curve has maximums at both \(t=0\) and \(t=0.94\). What is the frequency in Hz? The tolerance is \(\pm 0.01\) Hz.

Solution

Count how many periods occur between \(t=0\) and \(t=0.94\).

Find the period.

\[P = \frac{0.94 ~\mathrm{seconds}}{3~\mathrm{cycles}}\]

The frequency is the reciprocal of the period.

\[F ~=~ \frac{1}{P} ~=~ \frac{3~\mathrm{cycles}}{0.94 ~\mathrm{seconds}}\]

\[F ~\approx~ 3.19 ~\mathrm{Hz}\]
Question

A sinusoidal pure tone is generated by randomly choosing a frequency (\(F\)) in the following equation: \[y(t) ~=~ \cos(2\pi F t)\]

As shown above, the curve has maximums at both \(t=0\) and \(t=0.97\). What is the frequency in Hz? The tolerance is \(\pm 0.01\) Hz.

Solution

Count how many periods occur between \(t=0\) and \(t=0.97\).

Find the period.

\[P = \frac{0.97 ~\mathrm{seconds}}{7~\mathrm{cycles}}\]

The frequency is the reciprocal of the period.

\[F ~=~ \frac{1}{P} ~=~ \frac{7~\mathrm{cycles}}{0.97 ~\mathrm{seconds}}\]

\[F ~\approx~ 7.22 ~\mathrm{Hz}\]
Question

A sinusoidal pure tone is generated by randomly choosing a frequency (\(F\)) in the following equation: \[y(t) ~=~ \cos(2\pi F t)\]

As shown above, the curve has maximums at both \(t=0\) and \(t=0.97\). What is the frequency in Hz? The tolerance is \(\pm 0.01\) Hz.

Solution

Count how many periods occur between \(t=0\) and \(t=0.97\).

Find the period.

\[P = \frac{0.97 ~\mathrm{seconds}}{7~\mathrm{cycles}}\]

The frequency is the reciprocal of the period.

\[F ~=~ \frac{1}{P} ~=~ \frac{7~\mathrm{cycles}}{0.97 ~\mathrm{seconds}}\]

\[F ~\approx~ 7.22 ~\mathrm{Hz}\]
Question

A sinusoidal pure tone is generated by randomly choosing a frequency (\(F\)) in the following equation: \[y(t) ~=~ \cos(2\pi F t)\]

As shown above, the curve has maximums at both \(t=0\) and \(t=0.91\). What is the frequency in Hz? The tolerance is \(\pm 0.01\) Hz.

Solution

Count how many periods occur between \(t=0\) and \(t=0.91\).

Find the period.

\[P = \frac{0.91 ~\mathrm{seconds}}{8~\mathrm{cycles}}\]

The frequency is the reciprocal of the period.

\[F ~=~ \frac{1}{P} ~=~ \frac{8~\mathrm{cycles}}{0.91 ~\mathrm{seconds}}\]

\[F ~\approx~ 8.79 ~\mathrm{Hz}\]
Question

A sinusoidal pure tone is generated by randomly choosing a frequency (\(F\)) in the following equation: \[y(t) ~=~ \cos(2\pi F t)\]

As shown above, the curve has maximums at both \(t=0\) and \(t=0.91\). What is the frequency in Hz? The tolerance is \(\pm 0.01\) Hz.

Solution

Count how many periods occur between \(t=0\) and \(t=0.91\).

Find the period.

\[P = \frac{0.91 ~\mathrm{seconds}}{6~\mathrm{cycles}}\]

The frequency is the reciprocal of the period.

\[F ~=~ \frac{1}{P} ~=~ \frac{6~\mathrm{cycles}}{0.91 ~\mathrm{seconds}}\]

\[F ~\approx~ 6.59 ~\mathrm{Hz}\]
Question

A sinusoidal pure tone is generated by randomly choosing a frequency (\(F\)) in the following equation: \[y(t) ~=~ \cos(2\pi F t)\]

As shown above, the curve has maximums at both \(t=0\) and \(t=0.96\). What is the frequency in Hz? The tolerance is \(\pm 0.01\) Hz.

Solution

Count how many periods occur between \(t=0\) and \(t=0.96\).

Find the period.

\[P = \frac{0.96 ~\mathrm{seconds}}{7~\mathrm{cycles}}\]

The frequency is the reciprocal of the period.

\[F ~=~ \frac{1}{P} ~=~ \frac{7~\mathrm{cycles}}{0.96 ~\mathrm{seconds}}\]

\[F ~\approx~ 7.29 ~\mathrm{Hz}\]
Question

A sinusoidal pure tone is generated by randomly choosing a frequency (\(F\)) in the following equation: \[y(t) ~=~ \cos(2\pi F t)\]

As shown above, the curve has maximums at both \(t=0\) and \(t=0.97\). What is the frequency in Hz? The tolerance is \(\pm 0.01\) Hz.

Solution

Count how many periods occur between \(t=0\) and \(t=0.97\).

Find the period.

\[P = \frac{0.97 ~\mathrm{seconds}}{3~\mathrm{cycles}}\]

The frequency is the reciprocal of the period.

\[F ~=~ \frac{1}{P} ~=~ \frac{3~\mathrm{cycles}}{0.97 ~\mathrm{seconds}}\]

\[F ~\approx~ 3.09 ~\mathrm{Hz}\]
Question

A sinusoidal pure tone is generated by randomly choosing a frequency (\(F\)) in the following equation: \[y(t) ~=~ \cos(2\pi F t)\]

As shown above, the curve has maximums at both \(t=0\) and \(t=0.95\). What is the frequency in Hz? The tolerance is \(\pm 0.01\) Hz.

Solution

Count how many periods occur between \(t=0\) and \(t=0.95\).

Find the period.

\[P = \frac{0.95 ~\mathrm{seconds}}{7~\mathrm{cycles}}\]

The frequency is the reciprocal of the period.

\[F ~=~ \frac{1}{P} ~=~ \frac{7~\mathrm{cycles}}{0.95 ~\mathrm{seconds}}\]

\[F ~\approx~ 7.37 ~\mathrm{Hz}\]

Shared Qs (025)

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